?0,?2?x,?F(x)??2?3?1,?2x?1,?x?00?x?11?x?2x?2
27.求标准正态分布的上?分位点, (1)?=0.01,求z;
?(2)?=0.003,求z,z.
??/2【解】(1)
P(X?z?)?0.01
?即 1??(z即 ?(z故 (2) 由P(X?z?)?0.01
?)?0.09
z??2.33
)?0.003得
?1??(z?)?0.003即 ?(z)?0.997 查表得 由P(X?z?/2z??2.75
)?0.0015得
1??(z?/2)?0.0015
即 ?(z28.设随机变量X的分布律为 X ?/2)?0.9985查表得
z?/2?2.96?2 ?1 0 1 3 21
Pk 1/5 1/6 1/5 1/15 11/30 求Y=X2的分布律.
【解】Y可取的值为0,1,4,9
P(Y?0)?P(X?0)?15117??61530P(Y?1)?P(X??1)?P(X?1)?1P(Y?4)?P(X??2)?511P(Y?9)?P(X?3)?30
故Y的分布律为
Y Pk 0 1 4 9 1/5 7/30 1/5 11/30 k
29.设P{X=k}=(1), k=1,2,…,令 2?1,当X取偶数时Y????1,当X取奇数时.
求随机变量X的函数Y的分布律. 【解】P(Y?1)?P(X?2)?P(X?4)?L?P(X?2k)?L
111?()2?()4?L?()2k?L222111?()/(1?)?443
P(Y??1)?1?P(Y?1)?23
30.设X~N(0,1).
(1) 求Y=eX的概率密度;
22
(2) 求Y=2X2+1的概率密度; (3) 求Y=|X|的概率密度. 【解】(1) 当y≤0时,F(y)?P(Y?y)?0
Y当y>0时,F(y)?P(Y?y)?P(e?y)?P(X?lny) ??f(x)dx
xYlny??X故
fY(y)?dFY(y)111?ln2y/2?fx(lny)?e,y?0dyyy2π
(2)P(Y?2X2?1?1)?1
Y2当y≤1时F(y)?P(Y?y)?0 当y>1时F(y)?P(Y?y)?P(2XY?1?y)
?y?1?y?1??P?X2??P??X????22???y?1??2??
(y?1)/2?(y?1)/2 ??故
d1fY(y)?FY(y)?dy42??fXy?1??fX(x)dx
?y?1??y?1???fX??????????22??????
(3)
P(Y?0)?1?1221?(y?1)/4e,y?1y?12π
Y当y≤0时F(y)?P(Y?y)?0 当y>0时F(y)?P(|X|?y)?P(?y?X?y) ??f(x)dx
Yy?yX故f
Y(y)?dFY(y)?fX(y)?fX(?y)dy
23
?2?y2/2e,y?02π
31.设随机变量X~U(0,1),试求: (1) Y=eX的分布函数及密度函数; (2) Z=?2lnX的分布函数及密度函数. 【解】(1)
故
P(0?X?1)?1
XP(1?Y?eX?e)?1Y当y?1时F(y)?P(Y?y)?0 当1 ??dx?lny 当y≥e时F(y)?P(eYX?y)?1 即分布函数 y?1?0,?FY(y)??lny,1?y?e?1,y?e? 故Y的密度函数为 ?11?y?e?fY(y)??y,?0,其他? (2) 由P(0 P(Z?0)?1 当z≤0时,F(z)?P(Z?z)?0 Z当z>0时,F(z)?P(Z?z)?P(?2lnX?z) z)?P(X?e ?P(lnX??2Z?z/2) 24